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We started discussing recursive sequences in my Analysis I class last week (plus related questions 'is it monotonic?', 'does it converge?', 'is it bounded?' etc.). In my homework set I have the following problems:

(1) Show that the sequence given by

$$ \begin{aligned} a_1=2 \quad \quad \quad \quad \ \\ a_n=\frac{1}{2} \left(a_{n-1} + 6 \right) \end{aligned} $$

Is increasing and bounded above

And

(2) Consider the recursive sequence $a_{n+1} = 7-\frac{10}{a_n}$, with initial datum $a_1 = 4$. Compute the first two or three values. Show that it is bounded by 2 and 5, that it is increasing, and compute the limit.

During class, we discussed that the way to find whether a recursive sequence is increasing/decreasing is to compute the difference of $a_{n+1} - a_n$.

One of my classmates explained to me that this difference for (1) equals $3-\frac{a_n}{2}$, but he couldn't explain to me how he got to this value.

Concretely, my question is:

a) How do I compute the difference between $a_{n-1} - a_n$? What sort of reasoning should I employ to do this? To be honest, the subscripts mess me up quite a bit.

b) I had some trouble finding literature on this. Do you know of any resources where I can read up on this?

(Note: I copied the questions here for completeness, but I'm not asking you to help me to show that the sequences is bounded. I can reason through the induction proof, I'm just a bit clueless about this first step.)

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  • $\begingroup$ Hi! The value for the difference isn't exactly accurate - $a_{n+1}-a_n=3-\frac{a_n}2$ $\endgroup$
    – user418131
    Oct 12, 2017 at 12:24
  • $\begingroup$ To see why that is true, simply subtract $a_n$ from $a_{n+1}$. Then possibly conjecture, and then prove, that $\{a_n\}$ is increasing but bounded above by 6, using induction on $n$ $\endgroup$
    – user418131
    Oct 12, 2017 at 12:25
  • $\begingroup$ Thanks for your comment. It isn’t clear to me how I can subtract these two values from one another. That was my question! $\endgroup$
    – JHG90
    Oct 12, 2017 at 13:00
  • $\begingroup$ Ok, so we have $a_{n+1}=\frac 12a_n+3\Rightarrow a_{n+1}-a_n=\frac 12a_n+3-a_n\;$:) $\endgroup$
    – user418131
    Oct 12, 2017 at 13:01
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    $\begingroup$ .... that was painfully obvious. Thank you! $\endgroup$
    – JHG90
    Oct 12, 2017 at 13:25

1 Answer 1

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We have $a_n+1=\frac 12a_n+3⇒a_n+1−a_n=\frac12a_n+3−a_n$. Now possibly conjecture, and then prove, that ${a_n}$ is increasing but bounded above by 6, using induction on $n$

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